$1/PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$n_{Al}=\dfrac{2,7}{27}=0,1mol.$
$n_{HCl}=\dfrac{200.7,3\%}{36,5}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{1}<\dfrac{0,4}{3}$
$⇒HCl$ $dư.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$2/Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,1mol.$
$⇒m_{AlCl_3}=0,1.133,5=13,35g.$
$⇒m_{ddAlCl_3}=m_{Al}+m_{ddHCl}-m_{H_2}=2,7+200-(0,15.2)=202,4g.$
$⇒C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{ddAlCl_3}}.100\%=\dfrac{13,35}{202,4}.100\%=6,59\%$
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